Probability of Winning a Dice Game: Homework Statement and Solutions

[KEØM]

Homework Statement

Sarah and Thomas each roll a die. Whoever gets the higher number wins; if they both roll they same number, neither wins.

a. What is the probability that Thomas wins?

b. If Sarah, rolls a 3 what is the probability that she wins?

c. If Sarah rolls a 3 what is the probability that Thomas wins?

d. If Sarah wins what is the probability that Thomas rolled a 3?

e. If Sarah wins what is the probability that Sarah rolled a three?

Homework Equations

P(A|B) = P(A$$\cap$$B)/P(B)

Bayes' Rule

The Attempt at a Solution

For a, is it as simple as they both at the beginning of the game have an equally likely chance of winning so 1/2?

b) If she rolls a 3 then in order for her to win Thomas must roll a 1 or 2 so the probability that she wins is 1/3

c) If she rolls a three then in order for Thomas to win he must roll a 4,5 or 6 so the probability that he wins is 1/2.

d) If Sarah wins she must have rolled a 2,3,4,5, or 6. For Thomas to lose he must have rolled a 1,2,3,4 or 5. So the probability that he rolls a 3 is 1/5.

e) If Sarah wins she must have rolled a 2,3,4,5, or 6. So the probability that she rolls a three is also 1/5.

Is this correct reasoning? Also how do I actually use the relevant equations in this problem?

 

[lanedance]

Homework Statement

Sarah and Thomas each roll a die. Whoever gets the higher number wins; if they both roll they same number, neither wins.

a. What is the probability that Thomas wins?

b. If Sarah, rolls a 3 what is the probability that she wins?

c. If Sarah rolls a 3 what is the probability that Thomas wins?

d. If Sarah wins what is the probability that Thomas rolled a 3?

e. If Sarah wins what is the probability that Sarah rolled a three?

Homework Equations

P(A|B) = P(A$$\cap$$B)/P(B)

Bayes' Rule

The Attempt at a Solution

For a, is it as simple as they both at the beginning of the game have an equally likely chance of winning so 1/2?

i don't think so, as there is also the potential no one wins...

b) If she rolls a 3 then in order for her to win Thomas must roll a 1 or 2 so the probability that she wins is 1/3

c) If she rolls a three then in order for Thomas to win he must roll a 4,5 or 6 so the probability that he wins is 1/2.

these sound right to me

d) If Sarah wins she must have rolled a 2,3,4,5, or 6. For Thomas to lose he must have rolled a 1,2,3,4 or 5. So the probability that he rolls a 3 is 1/5.

e) If Sarah wins she must have rolled a 2,3,4,5, or 6. So the probability that she rolls a three is also 1/5.

not sure i follow you reasoning on the last two

the question revolves around the random variable, X which can take values {1,2,..,6}
with P(X=n) = 1/6 for n in {1,2,..,6}

for example a) is given the two independent random variables X₁ & X₂ with distributions as given above, what is P(X₁ > X₂)...

however, as you were implying, due to symmetry, you know P(X₂ > X₁) = P(X₁ > X₂)

 

[CompuChip]

For a), I think not. If you write out all the possibilities for example, you will see that in less than half of them Thomas actually wins.

Similarly you over-simplified d) and e). Sarah winning does not mean that she rolled 2, 3, 4, 5 or 6 while Thomas got 1, 2, 3, 4 or 5. For example: if Sarah throws 3 and Thomas 4, then Thomas wins.